Question

3. Find the zeroes of the following quadratic polynomials and verify the
relationship between the zeroes and the coefficients.
(1) 4s² – 4s+1 (ii)
${6x}^{2} - 3 - 7x$
​

Answer 1

Answer:

सुभाषित पाठ के सभी श्लोकों को लिखें व याद करें ।

Answer 2

Step-by-step explanation:

$(i) \: 4 {s}^{2} - 4s + 1 \\ = 4 {s}^{2} - 2s - 2s+ 1 \\ = 2s(2s - 1) - 1(2s - 1) \\ = (2s - 1)(2s - 1) \\ s = \frac{1}{2} \: \: \: \: , \: \: \: \frac{1}{2} \\ \: \: \: \: \: \alpha \: \: \: \: , \: \: \: \: \beta \\ \blue{\mathfrak{⟨\underline{\large{sum \: of \: zeros}}}⟩} \\ \alpha + \beta = \frac{ - b}{a} \\ \frac{1}{2} + \frac{1}{2} = \frac{ - ( - 4)}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 = 1. \: verified \\ \blue{\mathfrak{⟨\underline{\large{product \: of \: zeros}}}⟩} \\ \alpha \beta = \frac{c}{a} \\ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{4} = \frac{1}{4}.verified \\(ii) \: 6 {x }^{2} - 3 - 7x \\ = 6 {x}^{2} - 7x - 3 \\ = 6 {x}^{2} - 9x + 2x - 3 \\ = 3x(2x - 3) + 1(2x - 3) \\ = (3x + 1)(2x - 3) \\ x = \frac{ - 1}{3} \: \: , \: \: \frac{3}{2} \\ \: \: \: \: \: \: \: \: \alpha \: \: \: \: , \: \: \ \beta \\ \blue{\mathfrak{⟨\underline{\large{sum \: of \: zeros}}}⟩} \\ \alpha + \beta = \frac{ - b}{a} \\ \frac{ - 1}{3} + \frac{3}{2} = \frac{ - ( - 7)}{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{7}{6} = \frac{7}{6} .verified \\ \blue{\mathfrak{⟨\underline{\large{product \: of \: zeros}}}⟩} \\ \alpha \beta = \frac{c}{a} \\ \frac{ - 1}{3} \times \frac{3}{2} = \frac{ - 3}{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{ - 1}{2} = \frac{ - 1}{2} .verified$

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